Ratio & Proportion

Express each of the following ratios in simplest form :

(i) 20 : 35

(ii) 95 : 57

(iii) 2.1 : 1.2

(iv) $3\dfrac{1}{4} : 6\dfrac{1}{2}$

(v) 6a : 9b

(vi) 8ab² : 6a²b

Answer

(i) 20 : 35

The given ratio is 20 : 35.

Let us find H.C.F. of 20 and 35:

$\begin{array}{r} 1 \\ 20 \overline{) 35} \\ \underline{-20} \\ 15 \overline{) 20} ( 1 \\ \underline{-15} \\ 5 \overline{) 15} ( 3 \\ \underline{-15} \\ 0 \end{array}$

H.C.F. = 5

$∴ 20 : 35 = \dfrac{20 \div 5}{35 \div 5} = \dfrac{4}{7} = 4 : 7$

Hence, the answer is 4 : 7

(ii) 95 : 57

The given ratio is 95 : 57.

Let us find H.C.F. of 95 and 57:

$\begin{array}{r} 1 \\ 57 \overline{) 95} \\ \underline{-57} \\ 38 \overline{) 57} ( 1 \\ \underline{-38} \\ 19 \overline{) 38} ( 2 \\ \underline{-38} \\ 0 \end{array}$

H.C.F. = 19

$∴ 95 : 57 = \dfrac{95 \div 19}{57 \div 19} = \dfrac{5}{3} = 5 : 3$

Hence, the answer is 5 : 3

(iii) 2.1 : 1.2

First, multiply both by 10 to remove decimals:

(2.1 x 10 : 1.2 x 10) = 21 : 12

Let us find H.C.F. of 21 and 12:

$\begin{array}{r} 1 \\ 12 \overline{) 21} \\ \underline{-12} \\ 9 \overline{) 12} ( 1 \\ \underline{-9} \\ 3 \overline{) 9} ( 3 \\ \underline{-9} \\ 0 \end{array}$

H.C.F. = 3

$∴ 21 : 12 = \dfrac{21 \div 3}{12 \div 3} = \dfrac{7}{4} = 7 : 4$

Hence, the answer is 7 : 4

(iv) $3\dfrac{1}{4} : 6\dfrac{1}{2}$

Convert mixed fractions to improper fractions: $\dfrac{13}{4} : \dfrac{13}{2}$

Let us find L.C.M. of 4 and 2:

$\begin{array}{l|rr} 2 & 4, & 2 \\ \hline 2 & 2, & 1 \\ \hline & 1, & 1 \end{array}$

L.C.M. = 2 x 2 = 4

Multiply both terms by 4 to convert fractions to whole number:

$\Big(\dfrac{13}{4} \times 4\Big) : \Big(\dfrac{13}{2} \times 4\Big) = 13 : 26$

Let us find H.C.F of 13 and 26:

$\begin{array}{r} 2 \\ 13 \overline{) 26} \\ \underline{-26} \\ 0 \end{array}$

H.C.F = 13

$\dfrac{13 \div 13}{26 \div 13}$ = 1 : 2

Hence, the answer is 1 : 2

(v) 6a : 9b

Given ratio is 6a : 9b

Let us find H.C.F. of 6 and 9:

$\begin{array}{r} 1 \\ 6 \overline{) 9} \\ \underline{-6} \\ 3 \overline{) 6} ( 2 \\ \underline{-6} \\ 0 \end{array}$

H.C.F. = 3

$\dfrac{6a}{9b} = \dfrac{6a \div 3}{9b \div 3} = \dfrac{2a}{3b} = 2a : 3b$

Hence, the answer is 2a : 3b

(vi) 8ab² : 6a²b

Find the common factors in the coefficients and variables:

Let us find H.C.F. of 8 and 6:

$\begin{array}{r} 1 \\ 6 \overline{) 8} \\ \underline{-6} \\ 2 \overline{) 6} ( 3 \\ \underline{-6} \\ 0 \end{array}$

H.C.F. = 2

Common variables: ab.

Divide both terms by 2ab:

$\dfrac{8ab^2}{2ab} : \dfrac{6a^2b}{2ab} = 4b : 3a$

Hence, the answer is 4b : 3a

Express each of the following ratios in simplest form :

(i) $\dfrac{1}{8} : \dfrac{1}{12}$

(ii) $\dfrac{1}{9} : \dfrac{1}{6}$

(iii) $\dfrac{2}{3} : \dfrac{5}{6}$

(iv) $\dfrac{1}{4} : \dfrac{1}{6} : \dfrac{1}{8}$

(v) $1\dfrac{2}{3} : 2\dfrac{1}{2} : 1\dfrac{3}{4}$

(vi) $2\dfrac{1}{3} : 3\dfrac{1}{4} : 1\dfrac{1}{6}$

Answer

(i) $\dfrac{1}{8} : \dfrac{1}{12}$

Let us find L.C.M of 8 and 12:

$\begin{array}{l|l} 2 & 8, 12 \\ \hline 2 & 4, 6 \\ \hline 2 & 2, 3 \\ \hline 3 & 1, 3 \\ \hline & 1, 1 \end{array}$

L.C.M. = 2 x 2 x 2 x 3 = 24.

Multiplying both terms by 24 to convert fractions to whole number:

$\left(\dfrac{1}{8} \times 24\right) : \left(\dfrac{1}{12} \times 24\right) = 3 : 2$

Hence, the answer is 3 : 2

(ii) $\dfrac{1}{9} : \dfrac{1}{6}$

Let us find L.C.M of 9 and 6:

$\begin{array}{l|l} 3 & 9, 6 \\ \hline 3 & 3, 2 \\ \hline 2 & 1, 2 \\ \hline & 1, 1 \end{array}$

L.C.M. = 3 x 3 x 2 = 18.

Multiplying both terms by 18 to convert fractions to whole number:

$\left(\dfrac{1}{9} \times 18\right) : \left(\dfrac{1}{6} \times 18\right) = 2 : 3$

Hence, the answer is 2 : 3

(iii) $\dfrac{2}{3} : \dfrac{5}{6}$

Let us find L.C.M of 3 and 6:

$\begin{array}{l|l} 3 & 3, 6 \\ \hline 2 & 1, 2 \\ \hline & 1, 1 \end{array}$

L.C.M. = 3 x 2 = 6.

Multiplying both terms by 6 to convert fractions to whole number:

$\left(\dfrac{2}{3} \times 6\right) : \left(\dfrac{5}{6} \times 6\right) = 4 : 5$

Hence, the answer is 4 : 5

(iv) $\dfrac{1}{4} : \dfrac{1}{6} : \dfrac{1}{8}$

Let us find L.C.M of 4, 6 and 8:

$\begin{array}{l|l} 2 & 4, 6, 8 \\ \hline 2 & 2, 3, 4 \\ \hline 2 & 1, 3, 2 \\ \hline 3 & 1, 3, 1 \\ \hline & 1, 1, 1 \end{array}$

L.C.M. = 2 x 2 x 2 x 3 = 24.

Multiplying each term by 24 to convert fractions to whole number:

$\left(\dfrac{1}{4} \times 24\right) : \left(\dfrac{1}{6} \times 24\right) : \left(\dfrac{1}{8} \times 24\right) = 6 : 4 : 3$

Hence, the answer is 6 : 4 : 3

(v) $1\dfrac{2}{3} : 2\dfrac{1}{2} : 1\dfrac{3}{4}$

Convert to improper fractions: $\dfrac{5}{3} : \dfrac{5}{2} : \dfrac{7}{4}$.

Let us find L.C.M of 3, 2 and 4:

$\begin{array}{l|l} 2 & 3, 2, 4 \\ \hline 2 & 3, 1, 2 \\ \hline 3 & 3, 1, 1 \\ \hline & 1, 1, 1 \end{array}$

L.C.M. = 2 x 2 x 3 = 12.

Multiplying each term by 12 to convert fractions to whole number:

$\left(\dfrac{5}{3} \times 12\right) : \left(\dfrac{5}{2} \times 12\right) : \left(\dfrac{7}{4} \times 12\right) = 20 : 30 : 21$

Hence, the answer is 20 : 30 : 21

(vi) $2\dfrac{1}{3} : 3\dfrac{1}{4} : 1\dfrac{1}{6}$

Convert to improper fractions: $\dfrac{7}{3} : \dfrac{13}{4} : \dfrac{7}{6}$.

Let us find L.C.M of 3, 4 and 6:

$\begin{array}{l|l} 2 & 3, 4, 6 \\ \hline 2 & 3, 2, 3 \\ \hline 3 & 3, 1, 3 \\ \hline & 1, 1, 1 \end{array}$

L.C.M. = 2 x 2 x 3 = 12.

Multiplying each term by 12 to convert fractions to whole number:

$\left(\dfrac{7}{3} \times 12\right) : \left(\dfrac{13}{4} \times 12\right) : \left(\dfrac{7}{6} \times 12\right) = 28 : 39 : 14$

Hence, the answer is 28 : 39 : 14

Express each of the following ratios in simplest form :

(i) 35 paise : ₹ 1

(ii) 1 kg : 375 g

(iii) 60 cm : 1 m

(iv) 400 m : 1 km

(v) 8 months : $1\dfrac{2}{3}$ years

(vi) $6\dfrac{2}{3}$ hours : 1 day

Answer

(i) 35 paise : ₹ 1

Convert rupee into paise: ₹ 1 = 100 paise.

The ratio is 35 : 100.

Let us find H.C.F. of 35 and 100:

$\begin{array}{r} 2 \\ 35 \overline{) 100} \\ \underline{-70} \\ 30 \overline{) 35} ( 1 \\ \underline{-30} \\ 5 \overline{) 30} ( 6 \\ \underline{-30} \\ 0 \end{array}$

H.C.F. = 5

$\dfrac{35 \div 5}{100 \div 5} = \dfrac{7}{20} = 7 : 20$

Hence, the answer is 7 : 20

(ii) 1 kg : 375 g

Convert kg into g: 1 kg = 1000 g.

The ratio is 1000 : 375.

Let us find H.C.F. of 1000 and 375:

$\begin{array}{r} 2 \\ 375 \overline{) 1000} \\ \underline{-750} \\ 250 \overline{) 375} ( 1 \\ \underline{-250} \\ 125 \overline{) 250} ( 2 \\ \underline{-250} \\ 0 \end{array}$

H.C.F. = 125

$\dfrac{1000 \div 125}{375 \div 125} = \dfrac{8}{3} = 8 : 3$

Hence, the answer is 8 : 3

(iii) 60 cm : 1 m

Convert m into cm: 1 m = 100 cm.

The ratio is 60 : 100.

Let us find H.C.F. of 60 and 100:

$\begin{array}{r} 1 \\ 60 \overline{) 100} \\ \underline{-60} \\ 40 \overline{) 60} ( 1 \\ \underline{-40} \\ 20 \overline{) 40} ( 2 \\ \underline{-40} \\ 0 \end{array}$

H.C.F. = 20

$\dfrac{60 \div 20}{100 \div 20} = \dfrac{3}{5} = 3 : 5$

Hence, the answer is 3 : 5

(iv) 400 m : 1 km

Convert km into m: 1 km = 1000 m.

The ratio is 400 : 1000.

Let us find H.C.F. of 400 and 1000:

$\begin{array}{r} 2 \\ 400 \overline{) 1000} \\ \underline{-800} \\ 200 \overline{) 400} ( 2 \\ \underline{-400} \\ 0 \end{array}$

H.C.F. = 200

$\dfrac{400 \div 200}{1000 \div 200} = \dfrac{2}{5} = 2 : 5$

Hence, the answer is 2 : 5

(v) 8 months : $1\dfrac{2}{3}$ years

Convert $1\dfrac{2}{3}$ years to improper fraction: $\dfrac{5}{3}$ years.

Convert years into months: $\dfrac{5}{3} \times 12 = 20$ months.

The ratio is 8 : 20.

Let us find H.C.F. of 8 and 20:

$\begin{array}{r} 2 \\ 8 \overline{) 20} \\ \underline{-16} \\ 4 \overline{) 8} ( 2 \\ \underline{-8} \\ 0 \end{array}$

H.C.F. = 4

$\dfrac{8 \div 4}{20 \div 4} = \dfrac{2}{5} = 2 : 5$

Hence, the answer is 2 : 5

(vi) $6\dfrac{2}{3}$ hours : 1 day

Convert $6\dfrac{2}{3}$ hours to improper fraction: $\dfrac{20}{3}$ hours.

Convert 1 day into hours: 24 hours.

The ratio is $\dfrac{20}{3}$ : 24.

Multiply both sides by 3 to remove the denominator:

$\Big(\dfrac{20}{3} \times 3\Big) : 24 \times 3 = 20 : 72$

Let us find H.C.F. of 20 and 72:

$\begin{array}{r} 3 \\ 20 \overline{) 72} \\ \underline{-60} \\ 12 \overline{) 20} ( 1 \\ \underline{-12} \\ 8 \overline{) 12} ( 1 \\ \underline{-8} \\ 4 \overline{) 8} ( 2 \\ \underline{-8} \\ 0 \end{array}$

H.C.F. = 4

$\dfrac{20 \div 4}{72 \div 4} = \dfrac{5}{18} = 5 : 18$

Hence, the answer is 5 : 18

The angles of a triangle are in the ratio 5 : 6 : 7. Find each angle.

Answer

Given:

Ratio = 5 : 6 : 7.

The sum of all angles in any triangle is always 180°.

Find total parts: 5 + 6 + 7 = 18 parts.

Find the value of 1 part: 180° ÷ 18 = 10°

Now, calculate each angle by multiplying each part with 10°:

First angle: 5 x 10° = 50°

Second angle: 6 x 10° = 60°

Third angle: 7 x 10° = 70°

Angles are 50°, 60°, and 70°.

A line segment of length 84 cm has been divided into two parts in the ratio 4 : 3. Find the length of each part.

Answer

Given:

Length = 84 cm

Ratio = 4 : 3

Find total parts: 4 + 3 = 7 parts.

Find the value of 1 part: 84 ÷ 7 = 12 cm.

Now, calculate each part by multiplying with 12 cm:

First part = 4 x 12 cm = 48 cm

Second part = 3 x 12 cm = 36 cm

Lengths are 48 cm and 36 cm.

Divide ₹ 3900 between Kunal and Kirti in the ratio 15 : 11.

Answer

Given:

Total money = ₹ 3900

Ratio = 15 : 11

Find total parts: 15 + 11 = 26 parts.

Find the value of 1 part: Divide ₹ 3900 by 26.

3900 ÷ 26 = 150. So, 1 part = ₹ 150.

Now calculate shares by multiplying each part with ₹ 150:

Kunal's share: 15 x ₹ 150 = ₹ 2250

Kirti's share: 11 x ₹ 150 = ₹ 1650

Kunal's share = ₹2250 and Kirti's share = ₹1650.

Divide ₹ 16250 among A, B, C in the ratio 5 : 7 : 13.

Answer

Given:

Total Money: ₹ 16250

Ratio: 5 : 7 : 13

Find total parts: 5 + 7 + 13 = 25 parts.

Find the value of 1 part: 16250 ÷ 25 = 650. So, 1 part = ₹ 650.

Now calculate shares by multiplying each part with ₹ 650:

A gets: 5 x ₹ 650 = ₹ 3250

B gets: 7 x ₹ 650 = ₹ 4550

C gets: 13 x ₹ 650 = ₹ 8450

A's share = ₹3250, B's share = ₹4550, C's share = ₹8450

Divide 35 sweets between Sudha and Tanvy in the ratio $\dfrac{1}{2} : \dfrac{1}{3}$.

Answer

Given:

Total Sweets: 35

Ratio = $\dfrac{1}{2} : \dfrac{1}{3}$

Let us find L.C.M. of 2 and 3.

$\begin{array}{l|rr} 2 & 2, & 3 \\ \hline 3 & 1, & 3 \\ \hline & 1, & 1 \end{array}$

L.C.M. = 2 x 3 = 6

Multiply both by 6 to convert fractions to whole numbers:

$\Big(\dfrac{1}{2} \times 6\Big) : \Big(\dfrac{1}{3} \times 6\Big) = 3 : 2$

Total parts: 3 + 2 = 5 parts.

Value of 1 part: 35 ÷ 5 = 7 sweets.

Calculate sweets by multiplying each part with 7 sweets:

Sudha gets: 3 x 7 sweets = 21 sweets

Tanvy gets: 2 x 7 sweets = 14 sweets

Sudha gets 21 sweets and Tanvy gets 14 sweets.

Divide 468 g of rice into three heaps containing the quantities in the ratio $\dfrac{1}{6} : \dfrac{1}{8} : \dfrac{1}{12}$.

Answer

Given:

Total Rice: 468 g

Ratio: $\dfrac{1}{6} : \dfrac{1}{8} : \dfrac{1}{12}$

Let us find L.C.M. of 6, 8, and 12:

$\begin{array}{l|rrr} 2 & 6, & 8, & 12 \\ \hline 2 & 3, & 4, & 6 \\ \hline 2 & 3, & 2, & 3 \\ \hline 3 & 3, & 1, & 3 \\ \hline & 1, & 1, & 1 \end{array}$

L.C.M. = 2 x 2 x 2 x 3 = 24

Multiply all three terms with 24 to convert fractions to whole numbers:

$\Big(\dfrac{1}{6} \times 24\Big) : \Big(\dfrac{1}{8} \times 24\Big) : \Big(\dfrac{1}{12} \times 24\Big)$

= 4 : 3 : 2

Total parts: 4 + 3 + 2 = 9 parts.

Value of 1 part: 468 g ÷ 9 = 52 g.

Calculate heaps by multiplying each part with 52 g:

First heap: 4 x 52 g = 208 g

Second heap: 3 x 52 g = 156 g

Third heap: 2 x 52 g = 104 g

The three heaps are 208 g, 156 g and 104 g.

Divide ₹ 715 among A, B, C in such a way that B gets three times as much as A gets and C gets half of as much as B gets.

Answer

Given:

Total Money: ₹ 715

B gets 3 times A

C gets half of B.

Let A's share = 1 part.

Then B's share = 3 times A = 3 x 1 = 3 parts.

Then C's share = Half of B = 3 ÷ 2 = $\dfrac{3}{2}$ parts.

Ratio (A : B : C) = 1 : 3 : $\dfrac{3}{2}$

Multiply by 2, to convert fraction into whole number = 2 : 6 : 3.

Total parts: 2 + 6 + 3 = 11 parts.

Value of 1 part: 715 ÷ 11 = ₹ 65.

Calculate shares by multiplying each part with ₹ 65:

A gets: 2 x ₹ 65 = ₹ 130

B gets: 6 x ₹ 65 = ₹ 390

C gets: 3 x ₹ 65 = ₹ 195

A's share = ₹130, B's share = ₹390, C's share = ₹195.

Divide ₹ 760 among A, B, C such that A gets $\dfrac{5}{6}$ of what B gets and the ratio between the shares of B and C is 3 : 4.

Answer

Given:

Total Money: ₹ 760

A = $\dfrac{5}{6}$ of B

B : C = 3 : 4.

Since B : C is 3 : 4, let's look at A.

A = $\dfrac{5}{6} \times 3 = \dfrac{5}{2}$ parts.

Ratio (A : B : C) = $\dfrac{5}{2}$ : 3 : 4.

Multiply by 2 to convert fraction into whole number = 5 : 6 : 8.

Total parts: 5 + 6 + 8 = 19 parts.

Value of 1 part: 760 ÷ 19 = ₹ 40.

Calculate shares by multiplying each part with ₹ 40:

A gets: 5 x ₹ 40 = ₹ 200

B gets: 6 x ₹ 40 = ₹ 240

C gets: 8 x ₹ 40 = ₹ 320

A's share = ₹200, B's share = ₹240, C's share = ₹320

The boys and girls in a school are in the ratio 8 : 3. If the number of girls is 405, how many boys are there in the school?

Answer

Given:

Ratio (Boys : Girls) = 8 : 3

Total number of Girls = 405

Since 3 parts represent the girls, 3 parts = 405.

1 part = 405 ÷ 3 = 135.

Boys = 8 parts = 8 x 135 = 1080.

There are 1080 boys.

An alloy contains copper and zinc in the ratio 7 : 3. If it contains 12.6 g of copper, how much does this alloy weigh?

Answer

Given:

Ratio (Copper : Zinc) = 7 : 3

Copper = 12.6 g

Since 7 parts represent copper, 7 parts = 12.6 g

1 part = 12.6 ÷ 7 = 1.8 g

Zinc = 3 x 1.8 g = 5.4 g

Total weight = Copper + Zinc

Substituting the values in above:

Total weight = 12.6 g + 5.4 g = 18 g

The weight of the alloy is 18 g.

Reena weighed 63 kg. She reduced her weight in the ratio 9 : 8. Find her new weight.

Answer

Given:

Original Weight = 63 kg

Reduction Ratio = Old weight : New weight = 9 : 8

9 parts = 63 kg

1 part = 63 ÷ 9 = 7 kg.

The new weight is represented by 8 parts.

New weight 8 x 7 kg = 56 kg.

Her new weight is 56 kg.

Two numbers are in the ratio 9 : 13 and their sum is 176. Find the numbers.

Answer

Given:

Ratio = 9 : 13

Total Sum = 176

Total parts = 9 + 13 = 22 parts.

Value of 1 part: 176 ÷ 22 = 8. (So, 1 part = 8).

Calculate the numbers by multiplying each part with 8:

First number = 9 x 8 = 72

Second number = 13 x 8 = 104

The numbers are 72 and 104.

Two numbers are in the ratio 5 : 8 and their difference is 12. Find the numbers.

Answer

Given:

Ratio = 5 : 8

Difference = 12

The larger part is 8 and the smaller is 5.

Difference in parts = 8 - 5 = 3 parts.

Difference given = 12

3 parts = 12

1 part = 12 ÷ 3 = 4

Calculate the numbers by multiplying each part with 4:

First number = 5 x 4 = 20

Second number = 8 x 4 = 32

The numbers are 20 and 32.

If a : b = 3 : 2 and b : c = 4 : 5, find a : c.

Answer

Given:

a : b = 3 : 2

b : c = 4 : 5

$a : b = \dfrac{a}{b} = \dfrac{3}{2} \\[1em] b : c = \dfrac{b}{c} = \dfrac{4}{5}$

Now,

$a : c = \dfrac{a}{c} = \Big( \dfrac{a}{b} \times \dfrac{b}{c} \Big) \\[1em] = \Big( \dfrac{3}{2} \times \dfrac{4}{5} \Big) \\[1em] = \Big( \dfrac{3}{1} \times \dfrac{2}{5} \Big) \\[1em] = \dfrac{6}{5} \\[1em] = 6 : 5$

∴ a : c = 6 : 5

If a : b = 5 : 6 and b : c = 2.8 : 3.5, find a : c.

Answer

Given:

a : b = 5 : 6

b : c = 2.8 : 3.5

Multiply by 10 to remove decimals in the second ratio.

b : c = (2.8 x 10) : (3.5 x 10) = 28 : 35

b : c = 4 : 5 $\quad$ [Dividing by 7]

$a : b = \dfrac{a}{b} = \dfrac{5}{6} \\[1em] b : c = \dfrac{b}{c} = \dfrac{4}{5}$

Now,

$a : c = \dfrac{a}{c} = \left( \dfrac{a}{b} \times \frac{b}{c} \right) \\[1em] = \left( \dfrac{5}{6} \times \dfrac{4}{5} \right) \\[1em] = \left( \dfrac{1}{6} \times \dfrac{4}{1} \right) \\[1em] = \dfrac{4}{6} \\[1em] = \dfrac{2}{3} \\[1em] = 2 : 3$

∴ a : c = 2 : 3

If p : q = $1\dfrac{1}{3} : 1\dfrac{1}{2}$ and q : r = $\dfrac{1}{2} : \dfrac{1}{3}$, find p : r.

Answer

Given:

p : q = $1\dfrac{1}{3} : 1\dfrac{1}{2}$

q : r = $\dfrac{1}{2} : \dfrac{1}{3}$

Convert p : q to improper fractions:

p : q = $\dfrac{4}{3} : \dfrac{3}{2}$

L.C.M. of 3 and 2 is 6.

Multiply p : q and q : r with 6 to convert fractions to whole numbers:

$p : q = (\dfrac{4}{3} \times 6) : (\dfrac{3}{2} \times 6) = 8 : 9$

And

$q : r = (\dfrac{1}{2} \times 6) : (\dfrac{1}{3} \times 6) = 3 : 2$

$\dfrac{p}{q} = \dfrac{8}{9} \\[1em] \dfrac{q}{r} = \dfrac{3}{2}$

Now,

$p : r = \dfrac{p}{r} = \left( \dfrac{p}{q} \times \dfrac{q}{r} \right) \\[1em] = \left( \dfrac{8}{9} \times \dfrac{3}{2} \right) \\[1em] = \left( \dfrac{8}{3} \times \dfrac{1}{2} \right) \\[1em] = \dfrac{8}{6} \\[1em] = \dfrac{4}{3} \\[1em] = 4 : 3$

∴ p : r = 4 : 3

If a : b = 3 : 4 and b : c = 8 : 9, find a : b : c.

Answer

Given:

a : b = 3 : 4

b : c = 8 : 9

We make b, the same in both the ratios.

Let us find L.C.M. of 4 and 8:

$\begin{array}{l|rr} 2 & 4, & 8 \\ \hline 2 & 2, & 4 \\ \hline 2 & 1, & 2 \\ \hline & 1, & 1 \end{array}$

L.C.M. = 2 x 2 x 2 = 8

So, we make b equal to 8 in each case.

Now, multiply the first ratio by 2:

$a : b = 3 : 4 \\[1em] = \dfrac{3 \times 2}{4 \times 2} \\[1em] = \dfrac{6}{8} \\[1em] = 6 : 8 \\[1em]$

In second ratio b has 8 in denominator already, so keep it as it is.

b : c = 8 : 9

∴ a : b : c = 6 : 8 : 9

If l : m = $2\dfrac{1}{2} : 1\dfrac{2}{3}$ and m : n = $1\dfrac{1}{4} : 3\dfrac{1}{2}$, find l : m : n.

Answer

Given:

l : m = $2\dfrac{1}{2} : 1\dfrac{2}{3}$

m : n = $1\dfrac{1}{4} : 3\dfrac{1}{2}$

Convert mixed to improper fractions:

l : m = $\dfrac{5}{2} : \dfrac{5}{3}$

m : n = $\dfrac{5}{4} : \dfrac{7}{2}$

Let us simplify the ratios into whole numbers.

l : m = $\dfrac{5}{2} : \dfrac{5}{3}$

Let us find L.C.M. of 2 and 3:

$\begin{array}{l|rr} 2 & 2, & 3 \\ \hline 3 & 1, & 3 \\ \hline & 1, & 1 \end{array}$

L.C.M. = 2 x 3 = 6

l : m = $\Big(\dfrac{5}{2} \times 6\Big) : \Big(\dfrac{5}{3} \times 6\Big)$ = 15 : 10

l : m = 3 : 2 $\quad$[Dividing by 5]

m : n = $\dfrac{5}{4} : \dfrac{7}{2}$

Let us find L.C.M. of 4 and 2:

$\begin{array}{l|rr} 2 & 4, & 2 \\ \hline 2 & 2, & 1 \\ \hline & 1, & 1 \end{array}$

L.C.M. = 2 x 2 = 4

m : n = $\Big(\dfrac{5}{4} \times 4\Big) : \Big(\dfrac{7}{2} \times 4\Big)$

m : n = 5 : 14

We make m, the same in both the ratios.

Let us find L.C.M. of 2 and 5:

$\begin{array}{l|rr} 2 & 2, & 5 \\ \hline 5 & 1, & 5 \\ \hline & 1, & 1 \end{array}$

L.C.M. = 2 x 5 = 10

So, we make m equal to 10 in each case.

Now, multiply the first ratio by 5:

$l : m = 3 : 2 \\[1em] = \dfrac{3 \times 5}{2 \times 5} \\[1em] = \dfrac{15}{10} \\[1em] = 15 : 10$

Multiply the second ratio by 2:

$m : n = 5 : 14 \\[1em] = \dfrac{5 \times 2}{14 \times 2} \\[1em] = \dfrac{10}{28} \\[1em] = 10 : 28$

Now we have,

l : m = 15 : 10 and m : n = 10 : 28

∴ l : m : n = 15 : 10 : 28

Which ratio is greater?

(i) (3 : 4) or (5 : 7)

The given ratios are 3 : 4 and 5 : 7, which are equivalent to the fractions $\dfrac{3}{4}$ and $\dfrac{5}{7}$ respectively.

Let us find the L.C.M. of 4 and 7:

$\begin{array}{l|rr} 2 & 4, & 7 \\ \hline 2 & 2, & 7 \\ \hline 7 & 1, & 7 \\ \hline & 1, & 1 \end{array}$

L.C.M. = 2 x 2 x 7 = 28.

$∴ 3 : 4 = \dfrac{3}{4} = \dfrac{3 \times 7}{4 \times 7} = \dfrac{21}{28}$

$5 : 7 = \dfrac{5}{7} = \dfrac{5 \times 4}{7 \times 4} = \dfrac{20}{28}$

Clearly, $\dfrac{21}{28} \gt \dfrac{20}{28}$.

Hence, (3 : 4) > (5 : 7).

(ii) (11 : 21) or (19 : 28)

Answer

The given ratios are 11 : 21 and 19 : 28, which are equivalent to the fractions $\dfrac{11}{21}$ and $\dfrac{19}{28}$ respectively.

Let us find L.C.M. of 21 and 28:

$\begin{array}{l|l} 7 & 21, 28 \\ \hline 3 & 3, 4 \\ \hline 4 & 1, 4 \\ \hline & 1, 1 \end{array}$

L.C.M. = 7 x 3 x 4 = 84.

$\therefore 11 : 21 = \dfrac{11}{21} = \dfrac{11 \times 4}{21 \times 4} = \dfrac{44}{84}$

$19 : 28 = \dfrac{19}{28} = \dfrac{19 \times 3}{28 \times 3} = \dfrac{57}{84}$

Clearly, $\dfrac{57}{84} \gt \dfrac{44}{84}$.

Hence, (19 : 28) > (11 : 21).

Which of the following statements are true?

(i) 51 : 68 = 85 : 102

(ii) 1.5 : 2.5 = 3.6 : 6

(iii) 30 bags : 18 bags = ₹ 450 : ₹ 270

(iv) 32 kg : ₹ 36 = 8 kg : ₹ 9

Answer

(i) False
Reason —

First Ratio (51 : 68):

Let us find H.C.F. of 51 and 68:

$\begin{array}{r} 1 \\ 51 \overline{) 68} \\ \underline{-51} \\ 17 \overline{) 51} ( 3 \\ \underline{-51} \\ 0 \end{array}$

H.C.F. is 17.

Simplest form: $\dfrac{51 \div 17}{68 \div 17} = {\dfrac{3}{4}}$

Second Ratio (85 : 102):

Let us find H.C.F. of 85 and 102:

$\begin{array}{r} 1 \\ 85 \overline{) 102} \\ \underline{-85} \\ 17 \overline{) 85} ( 5 \\ \underline{-85} \\ 0 \end{array}$

H.C.F. is 17.

Simplest form: $\dfrac{85 \div 17}{102 \div 17} = {\dfrac{5}{6}}$

Since $3:4 \neq 5:6$, the statement is False.

(ii) True
Reason —

First Ratio = 1.5 : 2.5

Multiply by 10 to remove decimals = (1.5 x 10 : 2.5 x 10) = 15 : 25

Let us find H.C.F. of 15 and 25:

$\begin{array}{r} 1 \\ 15 \overline{) 25} \\ \underline{-15} \\ 10 \overline{) 15} ( 1 \\ \underline{-10} \\ 5 \overline{) 10} ( 2 \\ \underline{-10} \\ 0 \end{array}$

H.C.F = 5

Simplest form: $\dfrac{15 \div 5}{25 \div 5} = {\dfrac{3}{5}}$

Second Ratio = 3.6 : 6

Multiply by 10 to remove decimals = (3.6 x 10 : 6 x 10) = 36 : 60

Let us find H.C.F. of 36 and 60:

$\begin{array}{r} 1 \\ 36 \overline{) 60} \\ \underline{-36} \\ 24 \overline{) 36} ( 1 \\ \underline{-24} \\ 12 \overline{) 24} ( 2 \\ \underline{-24} \\ 0 \end{array}$

H.C.F. = 12.

Simplest form: $\dfrac{36 \div 12}{60 \div 12} = {\dfrac{3}{5}}$

Since both simplify to 3:5, the statement is True.

(iii) True
Reason —

First Ratio = 30 : 18

Let us find H.C.F. of 30 and 18:

$\begin{array}{r} 1 \\ 18 \overline{) 30} \\ \underline{-18} \\ 12 \overline{) 18} ( 1 \\ \underline{-12} \\ 6 \overline{) 12} ( 2 \\ \underline{-12} \\ 0 \end{array}$

H.C.F. = 6.

Simplest form: $\dfrac{30 \div 6}{18 \div 6} = {\dfrac{5}{3}}$

Second Ratio = 450 : 270

First, cancel the zeros: 45 : 27.

Let us find H.C.F. of 45 and 27:

$\begin{array}{r} 1 \\ 27 \overline{) 45} \\ \underline{-27} \\ 18 \overline{) 27} ( 1 \\ \underline{-18} \\ 9 \overline{) 18} ( 2 \\ \underline{-18} \\ 0 \end{array}$

H.C.F. = 9.

Simplest form: $\dfrac{45 \div 9}{27 \div 9} = {\dfrac{5}{3}}$

Since both simplify to 5:3, the statement is True.

(iv) False
Reason —

First Ratio = 32 : 36

Let us find H.C.F. of 32 and 36:

$\begin{array}{r} 1 \\ 32 \overline{) 36} \\ \underline{-32} \\ 4 \overline{) 32} ( 8 \\ \underline{-32} \\ 0 \end{array}$

H.C.F. = 4.

Simplest form: $\dfrac{32 \div 4}{36 \div 4} = {\dfrac{8}{9}}$

Second Ratio = 8 : 9

This is already in its simplest form: ${\dfrac{8}{9}}$

So numerically both ratios are equal.

But a ratio should exist only between quantities of the same kind. Here the quantities are kg and ₹, which are different kinds. Therefore the statement is false.

Check whether the following numbers are in proportion or not :

(i) 30, 40, 45, 60

(ii) $2, 2\dfrac{1}{2}, 3, 3\dfrac{1}{2}$

(iii) 0.8, 3, 2.4, 9

(iv) $\dfrac{1}{5},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{10}$

(v) $\dfrac{1}{2},\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{15}$

Answer

(i) 30, 40, 45, 60

The given numbers are 30, 40, 45, 60.

We have:

$30 : 40 = \dfrac{30}{40} = \dfrac{3}{4}$

$45 : 60 = \dfrac{45}{60} = \dfrac{45 \div 15}{60 \div 15} = \dfrac{3}{4}$

∴ (30 : 40) = (45 : 60)

Hence, 30, 40, 45 and 60 are in proportion.

(ii) $2, 2\dfrac{1}{2}, 3, 3\dfrac{1}{2}$

The given numbers are $2, 2\dfrac{1}{2}, 3, 3\dfrac{1}{2}$.

Convert mixed to improper fraction:

$2\dfrac{1}{2} = \dfrac{5}{2}$

$3\dfrac{1}{2} = \dfrac{7}{2}$

We have:

$2 : \dfrac{5}{2} = \dfrac{2}{5/2} = 2 \times \dfrac{2}{5} = \dfrac{4}{5}$

$3 : \dfrac{7}{2} = \dfrac{3}{7/2} = 3 \times \dfrac{2}{7} = \dfrac{6}{7}$

$\therefore (2 : 2\dfrac{1}{2}) \neq (3 : 3\dfrac{1}{2})$.

Hence, $2, 2\dfrac{1}{2}, 3, 3\dfrac{1}{2}$ are not in proportion.

(iii) 0.8, 3, 2.4, 9

The given numbers are 0.8, 3, 2.4, 9.

Multiply by 10 to convert decimals to whole numbers:

We have:

$0.8 : 3 = \dfrac{0.8 \times 10}{3 \times 10} = \dfrac{8}{30} = \dfrac{4}{15}$

$2.4 : 9 = \dfrac{2.4 \times 10}{9 \times 10} = \dfrac{24}{90} = \dfrac{24 \div 6}{90 \div 6} = \dfrac{4}{15}$

∴ (0.8 : 3) = (2.4 : 9)

Hence, 0.8, 3, 2.4 and 9 are in proportion.

(iv) $\dfrac{1}{5},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{10}$

The given numbers are $\dfrac{1}{5},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{10}$.

We have:

$\dfrac{1}{5} : \dfrac{1}{8} = \dfrac{1}{5} \times \dfrac{8}{1} = \dfrac{8}{5}$

$\dfrac{1}{4} : \dfrac{1}{10} = \dfrac{1}{4} \times \dfrac{10}{1} = \dfrac{10}{4} = \dfrac{5}{2}$

$∴ \dfrac{1}{5} : \dfrac{1}{8} \neq \dfrac{1}{4} : \dfrac{1}{10}$

Hence, $\dfrac{1}{5},\dfrac{1}{8},\dfrac{1}{4},\dfrac{1}{10}$ are not in proportion.

(v) $\dfrac{1}{2},\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{15}$

The given numbers are $\dfrac{1}{2},\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{15}$.

We have:

$\dfrac{1}{2} : \dfrac{1}{5} = \dfrac{1}{2} \times \dfrac{5}{1} = \dfrac{5}{2}$

$\dfrac{1}{6} : \dfrac{1}{15} = \dfrac{1}{6} \times \dfrac{15}{1} = \dfrac{15}{6} = \dfrac{5}{2}$

$∴ \dfrac{1}{2} : \dfrac{1}{5} = \dfrac{1}{6} : \dfrac{1}{15}$.

Hence, $\dfrac{1}{2},\dfrac{1}{5},\dfrac{1}{6},\dfrac{1}{15}$ are in proportion.

Find the value of x in each of the following :

(i) 42 : 12 : : 7 : x

(ii) 1.8 : x : : 2.4 : 6.0

(iii) 6 : 0.8 : : x : 10

(iv) x : 1.6 : : 2.1 : 8.4

(v) $\dfrac{1}{9}$ : x : : $\dfrac{1}{3} : \dfrac{1}{4}$

(vi) 16 : x : : x : 25

Answer

(i) 42 : 12 : : 7 : x

In a proportion, we know that:

product of extremes = product of means

∴ 12 x 7 = 42 x x

⇒ 84 = 42x

⇒ x = $\dfrac{84}{42} = 2$

Hence, x = 2

(ii) 1.8 : x : : 2.4 : 6.0

In a proportion, we know that:

product of extremes = product of means

∴ x x 2.4 = 1.8 x 6.0

⇒ 2.4x = 10.8

⇒ x = $\dfrac{10.8}{2.4} = \dfrac{108}{24} = \dfrac{9}{2} = 4.5$

Hence, x = 4.5

(iii) 6 : 0.8 : : x : 10

In a proportion, we know that:

product of extremes = product of means

∴ 0.8 x x = 6 x 10

⇒ 0.8x = 60

⇒ x = $\dfrac{60}{0.8} = \dfrac{600}{8} = 75$.

Hence, x = 75

(iv) x : 1.6 : : 2.1 : 8.4

In a proportion, we know that:

product of extremes = product of means

∴ 1.6 x 2.1 = x x 8.4

⇒ 3.36 = 8.4x

⇒ x = $\dfrac{3.36}{8.4} = \dfrac{336}{840} = 0.4$

Hence, x = 0.4

(v) $\dfrac{1}{9}$ : x : : $\dfrac{1}{3} : \dfrac{1}{4}$

In a proportion, we know that:

product of extremes = product of means

$∴ x \times \dfrac{1}{3} = \dfrac{1}{9} \times \dfrac{1}{4}$

⇒ $\dfrac{x}{3} = \dfrac{1}{36}$

⇒ x = $\dfrac{3}{36} = \dfrac{1}{12}$.

Hence, x = $\dfrac{1}{12}$

(vi) 16 : x : : x : 25

In a proportion, we know that:

product of extremes = product of means

∴ x x x = 16 x 25

⇒ x² = 400

⇒ x = $\sqrt{400} = 20$

Hence, x = 20

Find the fourth proportional to :

(i) 4, 9, 32

(ii) 15, 6, 7

(iii) 0.6, 1.5, 3

(iv) $\dfrac{1}{3}, \dfrac{2}{5}, 6$

(v) $2\dfrac{1}{2}, 2\dfrac{6}{7}, 3\dfrac{1}{2}$

(vi) 3 hrs 12 min, 24 min, 1 m 68 cm

Answer

(i) 4, 9, 32

Let the fourth proportional be x. Then 4, 9, 32, x are in proportion.

product of extremes = product of means

4 × x = 9 × 32

⇒ x = $\dfrac{9 \times 32}{4}$

⇒ x = 9 × 8 = 72

Hence, the fourth proportional is 72.

(ii) 15, 6, 7

Let the fourth proportional be x. Then 15, 6, 7, x are in proportion.

product of extremes = product of means

15 × x = 6 × 7

⇒ x = $\dfrac{42}{15} = \dfrac{14}{5} = 2.8$

Hence, the fourth proportional is 2.8

(iii) 0.6, 1.5, 3

Let the fourth proportional be x. Then 0.6, 1.5, 3, x are in proportion.

product of extremes = product of means

0.6 × x = 1.5 × 3

⇒ x = $\dfrac{4.5}{0.6} = \dfrac{4.5 \times 10}{0.6 \times 10} = \dfrac{45}{6} = 7.5$

Hence, the fourth proportional is 7.5

(iv) $\dfrac{1}{3}, \dfrac{2}{5}, 6$

Let the fourth proportional be x. Then $\dfrac{1}{3}, \dfrac{2}{5}, 6$, x are in proportion.

product of extremes = product of means

$\dfrac{1}{3} \times x = \dfrac{2}{5} \times 6 \\[1em] \Rightarrow \dfrac{x}{3} = \dfrac{12}{5} \\[1em] \Rightarrow x = \dfrac{12 \times 3}{5} \\[1em] \Rightarrow x = \dfrac{36}{5} \\[1em] \Rightarrow x = 7\dfrac{1}{5}$

Hence, the fourth proportional is $7\dfrac{1}{5}$

(v) $2\dfrac{1}{2}, 2\dfrac{6}{7}, 3\dfrac{1}{2}$

Convert to improper fractions: $\dfrac{5}{2}, \dfrac{20}{7}, \dfrac{7}{2}$. Let the fourth proportional be x.

Then $\dfrac{5}{2}, \dfrac{20}{7}, \dfrac{7}{2}$, x are in proportion.

product of extremes = product of means

$\dfrac{5}{2} \times x = \dfrac{20}{7} \times \dfrac{7}{2} \\[1em] \Rightarrow \dfrac{5x}{2} = 10 \\[1em] \Rightarrow 5x = 10 \times 2 \\[1em] \Rightarrow 5x = 20 \\[1em] \Rightarrow x = \dfrac{20}{5} \\[1em] \Rightarrow x = 4$

Hence, the fourth proportional is 4

(vi) 3 hrs 12 min, 24 min, 1 m 68 cm

First, convert to the same units for each ratio:

1 hour = 60 min,

∴ 3 hrs = 3 x 60 min = 180 min

3 hrs 12 min = 180 min + 12 min = 192 min.

1 m = 100 cm,

∴ 1 m 68 cm = 100 cm + 68 cm = 168 cm.

Let the fourth proportional be x (in cm). We have:

192 : 24 :: 168 : x

Then 192 min, 24 min, 168 cm, x are in proportion.

product of extremes = product of means

192 x x = 24 x 168

⇒ x = $\dfrac{24 \times 168}{192}$

⇒ x = $\dfrac{1 \times 168}{8}$

⇒ x = 21

Hence, the fourth proportional is 21 cm.

Find the mean proportion between :

(i) 81 and 121

(ii) 1.8 and 0.2

(iii) $\dfrac{2}{3}$ and $\dfrac{8}{27}$

(iv) 0.32 and 0.08

(v) $\dfrac{1}{25}$ and $\dfrac{1}{16}$

Answer

(i) 81 and 121

Mean proportion between 81 and 121

$= \sqrt{81 \times 121} \\[1em] = \sqrt{81} \times \sqrt{121} \\[1em] = 9 \times 11 \\[1em] = 99 \\[1em]$

Hence, the answer is 99

(ii) 1.8 and 0.2

Mean proportion between 1.8 and 0.2

$= \sqrt{1.8 \times 0.2} \\[1em] = \sqrt{0.36} \\[1em] = 0.6$

Hence, the answer is 0.6

(iii) $\dfrac{2}{3}$ and $\dfrac{8}{27}$

Mean proportion between $\dfrac{2}{3}$ and $\dfrac{8}{27}$

$= \sqrt{\dfrac{2}{3} \times \dfrac{8}{27}} \\[1em] = \sqrt{\dfrac{16}{81}} \\[1em] = \dfrac{\sqrt{16}}{\sqrt{81}} \\[1em] = \dfrac{4}{9}$

Hence, the answer is $\dfrac{4}{9}$

(iv) 0.32 and 0.08

Mean proportion between 0.32 and 0.08

$= \sqrt{0.32 \times 0.08} \\[1em] = \sqrt{0.0256} \\[1em] = 0.16$

Hence, the answer is 0.16

(v) $\dfrac{1}{25}$ and $\dfrac{1}{16}$

Mean proportion between $\dfrac{1}{25}$ and $\dfrac{1}{16}$

$= \sqrt{\dfrac{1}{25} \times \dfrac{1}{16}} \\[1em] = \sqrt{\dfrac{1}{400}} \\[1em] = \dfrac{1}{\sqrt{400}} \\[1em] = \dfrac{1}{20}$

Hence, the answer is $\dfrac{1}{20}$

Find the third proportional to :

(i) 36, 12

(ii) 1.2, 0.6

(iii) $\dfrac{1}{9}, \dfrac{2}{3}$

(iv) 1m 60 cm, 40 cm

(v) 1 kg 250 g, 500 g

(vi) ₹ 2.40, ₹ 4.80

Answer

(i) 36, 12

Let the third proportional be x.

Then, 36 : 12 :: 12 : x.

product of extremes = product of means

36 × x = 12 × 12

⇒ x = $\dfrac{144}{36}$

⇒ x = 4

Hence, the third proportional is 4

(ii) 1.2, 0.6

Let the third proportional be x.

Then, 1.2 : 0.6 :: 0.6 : x.

product of extremes = product of means

1.2 × x = 0.6 × 0.6

⇒ x = $\dfrac{0.36}{1.2}$

⇒ x = $\dfrac{3.6}{12}$

⇒ x = 0.3

Hence, the third proportional is 0.3

(iii) $\dfrac{1}{9}, \dfrac{2}{3}$

Let the third proportional be x.

Then, $\dfrac{1}{9} : \dfrac{2}{3} :: \dfrac{2}{3} : x$.

product of extremes = product of means

$\dfrac{1}{9} \times x = \dfrac{2}{3} \times \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{x}{9} = \dfrac{4}{9} \\[1em] \Rightarrow x = \dfrac{4}{9} \times 9 \\[1em] \Rightarrow x = 4$

Hence, the third proportional is 4

(iv) 1 m 60 cm, 40 cm

First, convert to the same unit:

1 m = 100 cm

∴ 1 m 60 cm = 100 cm + 60 cm = 160 cm.

Let the third proportional be x.

Then, 160 : 40 :: 40 : x

product of extremes = product of means

160 × x = 40 × 40

⇒ x = $\dfrac{1600}{160}$

⇒ x = 10

Hence, the third proportional is 10 cm

(v) 1 kg 250 g, 500 g

First, convert to the same unit:

1 kg = 1000 g

∴ 1 kg 250 g = 1000 g + 250 g = 1250 g

Let the third proportional be x.

Then, 1250 : 500 :: 500 : x

product of extremes = product of means

1250 x x = 500 x 500

⇒ x = $\dfrac{250000}{1250}$

⇒ x = $\dfrac{25000}{125}$

⇒ x = 200

Hence, the third proportional is 200 g

(vi) ₹ 2.40, ₹ 4.80

Let the third proportional be x.

Then, 2.40 : 4.80 :: 4.80 : x.

product of extremes = product of means

2.40 × x = 4.80 × 4.80

⇒ x = $\dfrac{4.80 \times 4.80}{2.40}$

⇒ x = 2 x 4.80

⇒ x = 9.60

Hence, the third proportional is ₹ 9.60

Show that 6, 36, 216 are in continued proportion.

Answer

If 6, 36, 216 are in continued proportion, then it should satisfy the condition b² = a x c

Here a = 6, b = 36, c = 216

∴ 36² = 6 x 216

1296 = 1296

Since it satisfies the condition b² = a x c, the numbers 6, 36, and 216 are in continued proportion.

∴ 6, 36, 216 are in continued proportion

If 8 pens cost ₹ 356, what is the cost of 14 pens?

Answer

Given:

Cost of 8 pens = ₹ 356

Cost of 14 pens = ?

Let the cost of 14 pens = ₹ x

Ratio of pens : Ratio of costs

8 : 14 :: 356 : x

By the rule: Product of Extremes = Product of Means:

8 × x = 14 × 356

⇒ x = $\dfrac{14 \times 356}{8}$

⇒ x = 14 x 44.5

⇒ x = 623

Hence, the cost of 14 pens is ₹ 623.

A uniform iron bar of length 7 m weighs 22.4 kg. How much does the same bar of length 13 m weigh ?

Answer

Given:

Weight of 7 m iron bar = 22.4 kg

Weight of 13 m iron bar = ?

Let the weight of 13 m iron bar = x kg

Ratio of lengths : Ratio of weights

7 : 13 :: 22.4 : x

By the rule: Product of Extremes = Product of Means:

7 × x = 13 × 22.4

⇒ x = $\dfrac{13 \times 22.4}{7}$

⇒ x = 13 x 3.2

⇒ x = 41.6

Hence, the bar of length 13 m weighs 41.6 kg.

A distance of 68 km is represented on a map by 1.7 cm. What distance is represented by 8.5 cm on the same map?

Answer

Given:

Map distance 1.7 cm = Actual distance 68 km

Map distance 8.5 cm = ?

Let actual distance for 8.5 cm = x km

Ratio of map distances : Ratio of actual distances

1.7 : 8.5 :: 68 : x

By the rule: Product of Extremes = Product of Means:

1.7 x x = 8.5 x 68

⇒ x = $\dfrac{8.5 \times 68}{1.7}$

⇒ x = 5 x 68

⇒ x = 340

Hence, the distance represented is 340 km.

A bus is running at a uniform speed. It covers a distance of 435 km in 6 hours. How much distance will it cover in 8 hours ?

Answer

Given:

Distance covered in 6 hours = 435 km

Distance covered in 8 hours = ?

Let distance covered in 8 hours = x km

Ratio of times : Ratio of distances

6 : 8 :: 435 : x

By the rule: Product of Extremes = Product of Means:

6 × x = 8 × 435

⇒ x = $\dfrac{8 \times 435}{6}$

⇒ x = 8 x 72.5

⇒ x = 580

Hence, the bus will cover 580 km in 8 hours.

If 15 men can dig a trench 35 m long in 1 day, then how many men can dig a similar trench 84 m long in 1 day ?

Answer

Given:

Men required for 35 m trench = 15 men

Men required for 84 m trench = ?

Let men required for 84 m trench = x men

Length of trench : Number of men

35 : 84 :: 15 : x

By the rule: Product of Extremes = Product of Means:

35 × x = 84 × 15

x = $\dfrac{84 \times 15}{35}$

x = $\dfrac{84 \times 3}{7}$

x = 12 x 3

x = 36

Hence, 36 men are required to dig the 84 m trench.

The ratio between 6 cm and 20 mm is

1. 3 : 1
2. 3 : 10
3. 6 : 10
4. 3 : 2

Answer

Given:

First quantity = 6 cm

Second quantity = 20 mm

First, convert to the same unit (1 cm = 10 mm).

6 cm = 6 x 10 = 60 mm.

Ratio = 60 : 20 = $\dfrac{60}{20} = \dfrac{3}{1}$ = 3 : 1

Hence, option 1 is the correct option.

The ratio $\dfrac{1}{8} : \dfrac{1}{12}$ converted to the simplest form is

1. 12 : 8
2. 3 : 2
3. 2 : 5
4. 8 : 20

Answer

Given:

Ratio: $\dfrac{1}{8} : \dfrac{1}{12}$

Let us find the L.C.M. of denominators 8 and 12:

$\begin{array}{l|rr} 2 & 8, & 12 \\ \hline 2 & 4, & 6 \\ \hline 2 & 2, & 3 \\ \hline 3 & 1, & 3 \\ \hline & 1, & 1 \end{array}$

L.C.M. = 2 x 2 x 2 x 3 = 24

Multiply both terms by 24:

$\Big(\dfrac{1}{8} \times 24\Big) : \Big(\dfrac{1}{12} \times 24\Big) = 3 : 2$

Hence, option 2 is the correct option.

If A : B = 2 : 5 and B : C = 4 : 5, then C : A = ?

1. 5 : 2
2. 2 : 1
3. 15 : 8
4. 25 : 8

Answer

Given:

A : B = 2 : 5

B : C = 4 : 5

To find C : A, first find A : C by multiplying the ratios:

$\dfrac{A}{C} = \dfrac{A}{B} \times \dfrac{B}{C} \\[1em] = \dfrac{2}{5} \times \dfrac{4}{5} \\[1em] = \dfrac{8}{25}$

So, A : C = 8 : 25.

For C : A, we reverse the ratio = 25 : 8.

Hence, option 4 is the correct option.

By increasing 91 in the ratio 7 : 13, we get :

1. 182
2. 169
3. 121
4. 116

Answer

Given:

Original number = 91

Increased ratio = 7 : 13

The original part is 7.

7 parts = 91

1 part = 91 ÷ 7 = 13

The increased value is 13 parts:

New value = 13 x 13 = 169

Hence, option 2 is the correct option.

If 12 : x : : 15 : 25, then the value of x is

1. 10
2. 15
3. 18
4. 20

Answer

Given:

Proportion: 12 : x : : 15 : 25

Product of Means = Product of Extremes

x × 15 = 12 × 25

15x = 300

⇒ x = $\dfrac{300}{15}$

⇒ x = 20

Hence, option 4 is the correct option.

The third proportional to 9 and 18 is

1. 21
2. 24
3. 27
4. 36

Answer

Given:

Numbers = 9 and 18

Let the third proportional be x.

9 : 18 : : 18 : x

Product of Means = Product of Extremes

9 × x = 18 × 18

⇒ x = $\dfrac{18 \times 18}{9}$

⇒ x = 2 x 18

⇒ x = 36

Hence, option 4 is the correct option.

The mean proportional between 5 and 45 is

1. 10
2. 12
3. 15
4. 25

Answer

Given:

Numbers = 5 and 45

Mean Proportion = $\sqrt{a \times b}$

= $\sqrt{5 \times 45} = \sqrt{225}$

= 15

Hence, option 3 is the correct option.

Which of the following are in continued proportion?

1. 4, 8, 12
2. 5, 15, 25
3. 6, 36, 216
4. 9, 12, 18

Answer

Let us consider the given numbers as: a, b, c

Check each option using b² = a x c:

8² = 64, 4 × 12 = 48 (No)

15² = 225, 5 × 25 = 125 (No)

36² = 1296, 6 × 216 = 1296 (Yes)

12² = 144, 9 × 18 = 162 (No)

Hence, option 3 is the correct option.

Fill in the blanks :

(i) Ratio has ............... unit.

(ii) To convert a ratio a : b in its simplest form, we divide a and b by ............... of a and b.

(iii) If a : b : : b : c, then a, b, c are said to be in ............... proportion.

(iv) If a, b, c are in continued proportion, then c is called the ............... proportional to a and b.

(v) In a proportion, the first and fourth terms are called the ............... .

Answer

(i) Ratio has no unit.

(ii) To convert a ratio a : b in its simplest form, we divide a and b by H.C.F. of a and b.

(iii) If a : b : : b : c, then a, b, c are said to be in continued proportion.

(iv) If a, b, c are in continued proportion, then c is called the third proportional to a and b.

(v) In a proportion, the first and fourth terms are called the extremes.

Write true (T) or false (F) :

(i) If a, b, c, d are in proportion, then ac = bd.

(ii) If a : b : : c : d, then a, b, c, d are said to be in absolute proportion.

(iii) If a, b, c, are in continued proportion, then the mean proportion b = $\dfrac{a + c}{2}$.

(iv) If x is the third proportional to a, b, then a : b : : b : x.

(v) 1, 2, 3, 4, are in proportion.

Answer

(i) False
Reason — For a, b, c, d to be in proportion (a : b :: c : d), the rule is Product of Extremes = Product of Means. This means a x d = b x c, or ad = bc. The statement says ac = bd, which is incorrect.

(ii) False
Reason — When four terms are in the form a : b :: c : d, they are simply said to be in proportion. There is no standard mathematical term called "absolute proportion" used in this context.

(iii) False
Reason — If a, b, c are in continued proportion, then a : b :: b : c. This means b² = ac or $b = \sqrt{ac}$. The formula $\dfrac{a+c}{2}$ is for the arithmetic mean, not the mean proportional.

(iv) True
Reason — By definition, if x is the third proportional to a and b, then a, b, and x are in continued proportion. In this sequence, b is the mean (repeated) term.

(v) False
Reason — To check if 1, 2, 3, 4 are in proportion, we test if 1 x 4 = 2 x 3:

Product of Extremes (1 x 4) = 4

Product of Means (2 x 3) = 6

Since 4 ≠ 6, they are not in proportion.

Ram Nath sold one of his properties worth ₹ 38,00,000. He wished to divide this money between his two daughters Priya and Seema in the ratio 7 : 12. He sold another property for ₹ 60,00,000. He divided this money between Priya and Seema in the ratio $\dfrac{1}{5} : \dfrac{1}{7}$

(1) What amount did Priya receive from the sale of second property ?

1. ₹ 14,00,000
2. ₹ 24,00,000
3. ₹ 25,00,000
4. ₹ 35,00,000

(2) What amount did Seema receive from the sale of first property ?

1. ₹ 14,00,000
2. ₹ 24,00,000
3. ₹ 25,00,000
4. ₹ 49,00,000

(3) The difference between the total amounts received by Priya and Seema is :

1. ₹ 0
2. ₹ 1,00,000
3. ₹ 2,00,000
4. ₹ 5,00,000

(4) The ratio between the amounts received by Seema from the sale of the first and the second properties is :

1. 1 : 1
2. 12 : 7
3. 24 : 25
4. 14 : 35

Answer

(1) Given:

Value of second property = ₹ 60,00,000

Ratio (Priya : Seema) = $\dfrac{1}{5} : \dfrac{1}{7}$

Let us find L.C.M. of 5 and 7:

$\begin{array}{l|rr} 5 & 5, & 7 \\ \hline 7 & 1, & 7 \\ \hline & 1, & 1 \end{array}$

L.C.M. = 5 x 7 = 35

Priya : Seema = $(\dfrac{1}{5} \times 35) : (\dfrac{1}{7} \times 35)$

Priya : Seema = 7 : 5

Total parts = 7 + 5 = 12

Value of 1 part = ₹ 60,00,000 ÷ 12 = ₹ 5,00,000

Priya's amount = 7 parts x ₹ 5,00,000 = ₹ 35,00,000

Hence, option 4 is the correct option.

(2) Given:

Value of first property = ₹ 38,00,000

Ratio (Priya : Seema) = 7 : 12

Total parts = 7 + 12 = 19

Value of 1 part = ₹ 38,00,000 ÷ 19 = ₹ 2,00,000

Seema's amount = 12 parts x ₹ 2,00,000 = ₹ 24,00,000

Hence, option 2 is the correct option.

(3)

Calculate total for Priya.

From 1st property:

Ratio (Priya : Seema) = 7 : 12 $\quad$Given

Value of 1 part = ₹ 2,00,000 $\quad$[From previous step]

∴ 7 x 2,00,000 = ₹ 14,00,000

From 2nd property:

Priya's amount = ₹ 35,00,000 $\quad$[From step 1]

Total = ₹ 14,00,000 + ₹ 35,00,000 = ₹ 49,00,000

Calculate total for Seema.

From 1st property:

Seema's amount = ₹ 24,00,000 $\quad$[From step 2]

From 2nd property:

Ratio (Priya : Seema) = $\dfrac{1}{5} : \dfrac{1}{7}$

L.C.M. of 5 and 7 is 35.

Priya : Seema = $(\dfrac{1}{5} \times 35) : (\dfrac{1}{7} \times 35)$

Priya : Seema = 7 : 5

Value of 1 part = ₹ 5,00,000

∴ 5 x ₹ 5,00,000 = ₹ 25,00,000

Total = ₹ 24,00,000 + ₹ 25,00,000 = ₹ 49,00,000

Difference = Priya - Seema

= ₹ 49,00,000 - ₹ 49,00,000 = ₹ 0

Hence, option 1 is the correct option.

(4)

Seema's 1st amount = ₹ 24,00,000 $\quad$[From step 2]

Seema's 2nd amount = ₹ 25,00,000 $\quad$[From previous step]

Ratio = 24,00,000 : 25,00,000

Ratio = 24 : 25

Hence, option 3 is the correct option.

Ranjan Singh makes statues of brass. Brass is an alloy of copper and zinc. Ranjan uses two varieties of brass for different kinds of statues. Variety 1 contains copper and zinc mixed in the ratio 7 : 4 and variety 2 contains these metals in the ratio 5 : 3. Ranjan makes an elephant statue from variety 1 and a horse statue from variety 2. The elephant statue weighs 176 g and it is known that the brass used in the horse statue contains 135 g zinc.

(1) Find the quantity of copper present in the brass used to make the elephant statue.

1. 98 g
2. 112 g
3. 121 g
4. 132 g

(2) How much copper is contained in the brass used to make the horse statue ?

1. 165 g
2. 175 g
3. 205 g
4. 225 g

(3) How much zinc is contained in the brass used to make the two statues ?

1. 169 g
2. 179 g
3. 189 g
4. 199 g